Private Blotto: Viewpoint Competition with Polarized Agents
Kate Donahue, Jon Kleinberg
TL;DR
This paper introduces the Private Blotto game, analyzing how independent agents allocate effort across items with different outcome functions, revealing stability conditions and implications for social media and political competition.
Contribution
It characterizes the Nash stability of the Private Blotto game and compares the effects of median versus mean outcome functions on effort allocation.
Findings
Stable arrangements are rare for median outcomes with many agents but can have high misallocated effort.
Unstable arrangements for mean outcomes can occur with many agents, but stable ones have low effort misallocation.
Implications for social media and political systems are discussed based on the game analysis.
Abstract
Social media platforms are responsible for collecting and disseminating vast quantities of content. Recently, however, they have also begun enlisting users in helping annotate this content - for example, to provide context or label disinformation. However, users may act strategically, sometimes reflecting biases (e.g. political) about the "right" label. How can social media platforms design their systems to use human time most efficiently? Historically, competition over multiple items has been explored in the Colonel Blotto game setting (Borel, 1921). However, they were originally designed to model two centrally-controlled armies competing over zero-sum "items", a specific scenario with limited modern-day application. In this work, we propose and study the Private Blotto game, a variant with the key difference that individual agents act independently, without being coordinated by a…
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TopicsGame Theory and Applications · Experimental Behavioral Economics Studies · Game Theory and Voting Systems
Private Blotto: Viewpoint Competition with Polarized Agents
Kate Donahue Department of Computer Science, Cornell University
Jon Kleinberg Departments of Information and Computer Science, Cornell University
Abstract
Colonel Blotto games are one of the oldest settings in game theory, originally proposed over a century ago in [Borel(1921)]. However, they were originally designed to model two centrally-controlled armies competing over zero-sum “fronts”, a specific scenario with limited modern-day application. In this work, we propose and study Private Blotto games, a variant connected to crowdsourcing and social media. One key difference in Private Blotto is that individual agents act independently, without being coordinated by a central “Colonel”. This model naturally arises from scenarios such as activist groups competing over multiple issues, partisan fund-raisers competing over elections in multiple states, or politically-biased social media users labeling news articles as misinformation. In this work, we completely characterize the Nash Stability of the Private Blotto game. Specifically, we show that the outcome function has a critical impact on the outcome of the game: we study whether a front is won by majority rule (median outcome) or a smoother outcome taking into account all agents (mean outcome). We study how this impacts the amount of “misallocated effort”, or agents whose choices doesn’t influence the final outcome. In general, mean outcome ensures that, if a stable arrangement exists, agents are close to evenly spaced across fronts, minimizing misallocated effort. However, mean outcome functions also have chaotic patterns as to when stable arrangements do and do not exist. For median outcome, we exactly characterize when a stable arrangement exists, but show that this outcome function frequently results in extremely unbalanced allocation of agents across fronts.
1 Introduction
Over a century ago, before modern game theory was fully established, Émile Borel proposed a family of zero-sum games inspired by two armies competing over multiple battlefields [Borel(1921)].
Definition 1** (Colonel Blotto).**
Two players, and , are competing over different fronts. The players have units of effort at their disposal respectively. Each player wins a front if they allocate more effort to the front than their opponent does, and each player wishes to win as many fronts as possible. Are there Nash stable arrangements of effort over fronts, and if so, which are they?
The name “Colonel Blotto” derives from the fact that a colonel controls multiple individual soldiers, which they are able to allocate across the battlefields in order to serve their overall objective. The Colonel Blotto game has been the focus of extensive exploration, including variants that allow for battlefields to have different values, for effort to be allocated probabalistically, and for smoother utility functions [Golman and Page(2009), Hart(2008), Osorio(2013)].
The game has also found application in areas far removed from warfare [Merolla et al.(2005)]. For example, in national politics, two political parties must decide how to allocate campaign budgets across multiple contested states; in marketing competition, two competing national retail chains must decide where to locate individual stores across different metropolitan areas [Merolla et al.(2005)]. Such situations have the structure of a Colonel Blotto game, with the political parties or retail chains as the colonels, and the geographic locations as the front where the competition takes place. Despite the different contexts, these scenarios all share a fundamental structure, with two centralized entities organizing effort over multiple fronts.
Decentralized conflicts.
By contrast, modern-day conflicts often involve players that may share common goals, but are not organized by a single actor. Focusing on political competition, which will be one of our primary motivating domains, there are many agents interested in influencing voters in an election — activists, local organizers, political bloggers, everyday voters with large social media presences — and while they may be collectively backing one of two political parties, many of them are not being explicitly directed in their actions by these parties. (A similar type of decentralization can take place in warfare itself, with guerrilla warfare involving allied groups that nevertheless choose their actions separately.)
How might we model this type of political viewpoint competition? We could imagine that there is a large collection of agents, each of whom is interested in taking part in a conflict with “fronts.” Each agent controls only one unit of effort, and can choose to devote that effort to one of the fronts. There is no centralized “colonel” to direct the agents, but there is an underlying structure that determines which agents are more or less closely aligned. In particular, each agent has a type, which we can think of as a viewpoint, bias, or political position; this type is represented by a real number . After each agent chooses a front to participate in, the outcome of the conflict on each given front is determined by an outcome function that takes the multiset of types at that front and determines a real-valued outcome.
We can think of the fronts as issues, for example; each agent can devote effort to influencing public opinion on one of these issues, and the resulting value of public opinion on an issue is some aggregation of the viewpoints of the agents who devote effort to this issue. We will be particularly interested in two of the most natural outcome functions for aggregating viewpoints in this setting: the median (in which the outcome on a front is the median of the types there) and the mean (in which the outcome is the mean of the types). Agents want the outcomes on each front (even the ones where they don’t participate) to match their types; thus, each agent experiences a cost equal to a weighted average of the distances between the outcome on each front and the agent’s type. (The agent’s payoff is simply the negative of this cost.) Here the weights in this average can be viewed as representing the relativel importance of the different fronts.
We will refer to this type of game as Private Blotto; like Colonel Blotto, it involves conflict over multiple fronts, but it is fundamentally different because it is designed to model decentralized conflict where each individual agent makes their own choice about which front to participate in.111In the military, a private is an enlisted soldier at the base of the hierarchy. This reflects our setting, which views the individual soldiers as the strategic actors, rather than the coordinating colonel who commands the army. We summarize the discussion above in the following definition.
Definition 2** (Private Blotto).**
Multiple agents are competing as the players in a game over different “fronts”. Each agent can choose exactly one front to compete in, and each agent has a type corresponding to a real number. An outcome function (for example, the median or mean) determines the outcome value on each front. An agent’s cost is equal to a weighted average distance between the outcome on each front and the agent’s type.
For this class of games, we can ask a number of basic questions. Given a set of agents specified by their types, and a number of fronts, are there Nash stable arrangements of agents over fronts? And how does the choice of outcome function affect the existence of stable arrangements?
We will see that these questions are already rich and complex even when all agents belong to one of two types, just as Colonel Blotto games already exhibit rich structure with just two colonels. Thus, we will focus much of our attention on this two-type case, when there are real numbers and , and each agent has a type equal to either or . In this case, the median outcome function is simply majority rule: the type with more agents on a given front becomes the outcome for that front. The mean outcome for a front lies somewhere in the interval between and , at a weighted average of these two endpoints determined by the number of agents of each type at this front.
We now provide some more detail on settings that can be modeled by the Private Blotto game, and then we give an overview of our results.
1.1 Motivating examples
Our Private Blotto formulation finds applicability in numerous modern-day settings. Here, we will describe a few key application areas in more detail.
Political contests and issue-based activism has historically been an application area for Colonel Blotto, such as [Merolla et al.(2005)]. We argue that Private Blotto might even be a more natural fit for this setting. Here, the fronts might represent issues or political campaigns, while the agents might be activist groups or donors, which might share similar goals, but are unable (for logistical or legal reasons) to coordinate their actions. Differing types would reflect differing political leanings, which could reflect a continuum of preferences among groups.
Crowdsourcing on social media has become a growing area of societal and academic interest in recent years [Yasseri and Menczer(2021), Wojcik et al.(2022), Allen et al.(2022), Pröllochs(2022)]. Typically, this setting involves asking social media users to provide labels on different items, such as whether certain news articles are misinformation or not. In this setting, the fronts might be news articles or items that need to be labeled. Each agent would be a social media user choosing which front to label. Here, player types might reflect partisan bias, which has been shown to affect how users select which news articles some users choose to fact-check in Twitter’s Birdwatch tool [Wojcik et al.(2022), Allen et al.(2022), Pröllochs(2022)]). In this setting, our model explores dynamics in how voluntary, biased users might label multiple fronts.
Finally, Private Blotto can also be used in modeling military engagements, the most traditional application of Colonel Blotto games. In the Private Blotto formulation, we assume that each agent might be an individual soldier, guerrilla member, or other actor that is acting without coordination from some central organizer. In this way, Private Blotto might naturally model more modern types of asymmetric warfare conflicts. Agents on the same “side” militarily are of the same type. Agent types allow us to model interactions between three or more militaries or combatant groups, as it reflects potential alliances or more nuanced conflicts between different agents.
1.2 Overview of results
We are interested in which instances of Private Blotto admit stable arrangements, by which we mean a choice of front by each agent that forms a pure Nash equilibrium of the game: no agent has an incentive to unilaterally change fronts. We also prove results characterizing the properties of stable arrangements.
We will find it useful to divide our analysis of the model into two main cases, depending on whether the number of agents is smaller or larger than the number of fronts. In the motivation from political viewpoint competition, these cases correspond to different settings, each meaningful: a small number of fronts relative to the number of agents can correspond to a large collection of political actors dividing their influence over a small number of key issues, while a large number of fronts relative to the number of agents arises in the crowdsourced settings discussed above, where individuals are asked to label a large universe of news articles or social media posts for misinformation.
When the number of fronts is small compared to the number of agents, we focus on the setting in which there are two types and . Here, let be the number of agents of type , and be the number of agents of type . For both the median and the mean outcome functions, it is natural to ask: for which pairs must there always exist a stable arrangement, and for which pairs is there no stable arrangement? In the case of the median outcome function, we find that for every number of fronts , the set of pairs that fail to yield a stable arrangement is finite, and we can exactly characterize this set of pairs for each as a certain median-critical region in the - plane. In the case of the mean outcome function, we show computationally that the subset of the - plane admitting stable arrangements has a complex structure that seems to lack a closed-form description, even with just two fronts with equal weights; we also show that with two fronts, when a stable arrangement does exist, it must divide the agents in an almost-proportional fashion.
We prove additional results about the structure of stable arrangements — how the agents divide across fronts in these arrangements — through the lens of misallocated effort. If we adopt the normative principle that fronts (representing issues for public debate) should receive a number of agents proportional to their weights, then our result for the mean outcome function shows that there is very little misallocation of effort relative to this proportionality principle. In contrast, we find that for the median outcome function, there can be very high levels of misallocation, with stable arrangements in which large numbers of agents choose fronts with relatively low weight.
Finally, we consider the case in which there are more fronts than agents; intuitively, these are settings where the agents have the ability to spread out arbitrarily thinly across fronts if they so choose, and where some fronts will inevitably be left empty. Here we identify sufficient conditions for stable arrangements to exist, and characterize the smallest number of agents for which there are instances without stable arrangements. With more fronts than agents, our results for median and mean are more similar to each other than they were for small numbers of fronts; this connects intuitively with the fact that each front will tend to have relatively few agents, leading to less of a distinction between the behavior of the median and mean. One difference between the functions is that is the minimum number of agents for which there can be instances lacking stable arrangements for the mean outcome function — every instance with at most three agents has a stable outcome — whereas is the corresponding minimum for the median outcome function.
2 Related works
2.1 Colonel Blotto
Colonel Blotto games (first proposed in [Borel(1921)]) is a game theoretic model where two different players, and , compete to allocate effort across multiple “fronts” or “fronts”, which may vary in how much each player values them. Typically, an player wins a front if they exert more effort there, and one main question of interest is when Nash equilibria of this system exist. The literature on Colonel Blotto games is extremely broad, so will focus on a few of the most relevant papers. Recently, [Ahmadinejad et al.(2019)] included a polynomial time algorithm for computing equilibria of the standard Colonel Blotto game, as well as related zero-sum games.
First, we will highlight some of the most commonly-studied variants of Colonel Blotto games. [Golman and Page(2009)] proposes the “General Blotto” game, which generalizes Colonel Blotto to permit multiple player types which have smooth utility functions over fronts and over combinations of fronts. [Hart(2008)] proposes the “General Lotto” game where each player selects a probabilistic distribution of effort over each fronts and gets utility given by the probability that a randomly drawn level of their effort beats their opponents’ random draw. Separately, [Osorio(2013)] proposes the “Lottery Blotto” game, where players allocate effort deterministically, but the player that allocates greater effort to a front doesn’t win deterministically, but rather probabalistically. This formulation is related to Tullock Contests Success functions (originally proposed in [Tullock(2001)], also studied in [Osorio(2013)], [Skaperdas(1996)]) where two players are competing over contests where they win probabalistically related to their effort (similar to our mean outcome function). [Goyal and Vigier(2014)] similarly studies contest functions where fronts are connected in a network and an asymmetric “attacker” and “defender” are allocating resources across these fronts.
Next, we will highlight the papers that are closest to ours. [Schwartz et al.(2014a)] gives Nash stability results for the Colonel Blotto game where players vary in their strength (amount of resources) and fronts vary in their value, so long as there are at least three fronts with each value. [Kovenock and Roberson(2012)] studies a limited form of coalitions where exactly two players and may form an alliance before playing a common opponent . [Boix-Adserà et al.(2020)] proposes and studies the “multi-player Colonel Blotto game”, which extends the classical Colonel Blotto structure to more than 2 players. [Anbarcı et al.(2020)] studies a variant of Colonel Blotto with more than two competing forces, but where fronts are presented sequentially, rather than simultaneously. [Mazur(2017)] studies Nash equilibria of Colonel Blotto games with exactly 2 fronts, but where outcome functions are constrained to be a polynomial function of the difference of each type’s allocation across the fronts.
In general, Colonel Blotto games and their variants differ from ours in a few ways. First, rather than assuming the player can control multiple agents, we assume each agent acts independently (a private citizen as opposed to a soldier). Because any of these agents could “win”, this dramatically increases the number of potential outcomes. However, in our second main difference, we assume that agents have some degree of similarity in their goals: agent may be more closely aligned with than , for example. Finally, we study a more general class of settings than is typically studied in Colonel Blotto, allowing for arbitrary numbers and valuations of fronts, as well as more general notions of winning (all or nothing, as well as a more smooth fractional utility).
2.2 Crowdsourcing
The area of crowdsourcing has been studied experimentally and theoretically in a wide range of papers. Again, we will focus on summarizing those that are most closely related to ours. Some papers, such as [Hettiachchi et al.(2022)], [Zhang et al.(2017)] study how to assign different crowd workers to multiple tasks in order to maximize the expected accuracy of labels.
A more nascent branch of crowdsourcing considers the case where crowd workers may have agency over which fronts they choose to label. [Zhang and van der Schaar(2012)] studies reputation-based mechanisms to incentivize crowd workers to exert effort on fronts that they are assigned. Our model is especially relevant to fact-checking on social media sites that allow voluntary labels by (potentially biased) users, such as on Facebook, Wikipedia and Twitter [Yasseri and Menczer(2021)], especially Twitter’s BirdWatch tool [Wojcik et al.(2022)]. [Allen et al.(2022)][Pröllochs(2022)] study how partisan affiliation, among other facts, affects how users on Twitter choose which tweets to fact check. [Saeed et al.(2022)] compares the accuracy of labels produced by voluntary, biased crowdsource workers to expert labels.
Our paper differs from most crowdsourcing papers in how it allows crowdsource workers to act as voluntary, potentially biased agents with some agency over which fronts they choose to label. Our work also differs stylistically in that we focus primarily on Nash Equilibria of such systems, an area that has typically not been explored previously.
3 Model
In this section, we make our theoretical model more precise. We assume there are fronts that total agents are competing over. We allow each front to have different weights or values , which are common values to each agent. Each agent controls exactly 1 unit of effort: they may choose which front to compete in, but may not coordinate with other players. However, agents come in types. Two agents of the same type have perfectly aligned incentives: when present on the same front, they work towards the same outcome, and when on different fronts, two agents of the same type are interchangeable. Each type has a real-valued bias that describes how similar or dissimilar it is to other types. For example, an agent of type is closer to an agent of than to .
3.1 Outcome functions
Once agents are arrayed on a front, the outcome of the battle is governed by an outcome function . In this paper, we will focus on two types of outcome functions: median outcome and proportional outcome. Given a set of agents on front , the median outcome function returns the median of the biases . If there are an even number of players on a particular front, then the median function averages together the middle two biases. Note that for the traditional two-player Colonel Blotto game, the median outcome function is equivalent to “winner-take-all”, where whichever type dominates the front wins. For types, the median outcome means that the front’s outcome is equivalent to the median , for some . Every other agent would experience a cost given by : the magnitude of difference between their biases.
On the other hand, the mean outcome function returns the mean of the biases: . This models a scenario where the final outcome of the front depends on the distribution of agent biases, not solely the median agent. Again, every agent experiences cost given by their distance to the final outcome: .
3.2 Agent cost
Even though each agent only participates in a single front, they still have preferences over all of the fronts. An agent of type will weight the cost of each front according to weights satisfying . If a front has no agents on it (), then we assume each agent experiences a cost for leaving the front empty. This cost is independent of agent bias, but is influenced by the weight. We include this feature to model settings where agents may choose to leave a front empty (potentially to join a more heavily-weighted or contested front), but suffer some non-zero cost in doing so. We can write the total cost as:
[TABLE]
We will say that an arrangement of agents onto fronts is stable if it satisfies Nash stability (no agent can unilaterally decrease its cost):
Definition 3** (Nash stability).**
An arrangement of players on fronts is (Nash) stable in the Private Blotto game if no agent can reduce its cost by switching from competing in one front to begin competing in another. Written in notation, this corresponds to requiring that for every agent of type with on front and every other front ,
[TABLE]
Nash stability is a natural definition of stability to study in this case because agents are assumed to model self-interested actors with the ability to move between fronts in order to reduce their cost. However, Nash stable solutions are not (in general) guaranteed to exist. In this paper, our main focus will be characterizing possible Nash stable arrangements.
4 More agents than fronts:
In analyzing the Private Blotto game, we will find it helpful to divide our analysis into two main regimes: when there are more agents than fronts (this section), and when there are fewer agents than fronts (Section 5). The setting where there are more agents than fronts models the case of numerous small agents competing over a limited set of fronts. As related to the examples in Section 1.1, it might model the case where there is a small number of physical fronts in combat, for example, where the number of combatants is much larger. For political issues, it could represent the case where there is a relatively small subset of major divisive issues that multiple political actors are debating. In this section, we will make one assumption in our analysis:
Assumption 1**.**
For the setting, we assume agents come in types, and , with biases .
Traditionally, Colonel Blotto games have mainly focused on the two-player case, which is analogous to our two-type case in Private Blotto [Borel(1921)]. We choose to present the two-type case in this section because it already admits a surprising amount of richness and complexity, while still being relatively clean. We relax this assumption in our analysis of the case in Section 5. Exploring further relaxations of these assumptions would be an interesting avenue for further analysis of the Private Blotto game. Without loss of generality, we will always name the two types so that (there are more type than type players).
Lemma 1 shows why Assumption 1 is helpful. So long as it holds and given sufficiently high cost for leaving a front empty, agents’ incentives become independent of bias values .
Lemma 1**.**
Given Assumption 1, so long as
[TABLE]
then no agent wishes to leave a front empty. As a result, agent strategy becomes independent of biases and relies solely on the number of agents of each type on each front, .
Proofs for Lemma 1, as well as for other proofs in this paper, are found in Appendix A.
Given Lemma 1, all of our results for the remainder of the section hold for every possible value of biases. For the rest of this section, unless stated otherwise, we will assume that the preconditions of Lemma 1 hold, which ensures that no front will be left empty. This assumption is mainly made for cleanness of analysis: if it is relaxed, then the value of causes minor changes in the stable arrangements, primarily for small numbers of agents .
4.1 Median outcome
First, in this section we explore Private Blotto with with median outcome function. We can view this setting as exploring the plane, studying for which values of a Nash stable arrangement exists, as well as constructively producing an example of a stable arrangement. Our results will be a function of the total number of fronts, , the common weights on each of these fronts, as well as , the number of agents of types and respectively. (Recall that given Lemma 1 all of our results will be independent of the biases . )
The first question we will ask is whether the number of unstable “points” ( pairs) is infinite or finite. Lemma 2 addresses this question by showing that whenever there are more than twice as many agents as there are fronts, there always exists a stable arrangement. This directly implies that, holding the number of fronts constant, there must only be a finite number of unstable points.
Lemma 2**.**
If (or with even), then there always exists a stable arrangement.
Proof.
We will prove this result constructively by producing an algorithm that always arrives at a stable arrangement.
Informally, this algorithm works by putting 2 type players on every front, stopping once either a) front is reached, or b) all of the type players have been assigned, or c) there are 3 type players left (which are then all assigned to the current front). Then, the algorithm places at least 2 type players on each item, again stopping once a) front is reached, or b) all of the type players have been allocated, or c) there are 3 type players left. This is a stable arrangement because for each front, each type wins by at least 2, so no single agent acting alone can change the outcome. For a formal description, see Algorithm 1.
If and both are even, then this algorithm will put exactly 2 agents on each front. If and both are even, then this will put at least 2 agents on each front. If and both are odd, then fronts will have type agents and fronts will have type agents. In total, this covers fronts, as desired. ∎
Lemma 2 described conditions where a stable arrangement exists, but left open the question of whether it might be possible to be more specifically characterize when stable arrangements fail to exist, a question that we will address next. In order to do so, we will need a more specific definition:
Definition 4** (Median-critical region).**
A set of parameters is in the median-critical region if they satisfy:
[TABLE]
and symmetrically if the roles of are reversed.
Lemma 3 immediately shows the importance of this definition: any set of parameters within the median-critical region must always result in an unstable arrangement.
Lemma 3**.**
For any set of parameters within the median-critical region with median outcome function and cost satisfying Lemma 1, there is never a stable arrangement of agents onto fronts, no matter the weights.
Proof sketch.
To start out with, we will describe a few arrangements where a deviation is always possible. The second half of this proof will show that if we have any arrangement satisfying the preconditions (parameters within the median-critical region), then at least one of these cases must occur, meaning that the arrangement must be unstable.
For notation, we will use to denote an arrangement where there are at least as many type players as type on front 1, for example. We will only refer to fronts 1 and 2 for convenience, but these results hold for any fronts.
[TABLE]
This gives an arrangement where type loses by 1 on front 1 and wins by at least 2 on front 2. This gives a deviation because any player from front 2 could move to compete in front 1 and strictly reduce their cost (they now tie in the first and still win in the second). Similar reasoning holds for the other case: the type player from front 2 could move to compete in front 1 and strictly reduce their cost.
[TABLE]
Here, the players tie on front 1 and type wins by at least 2 on front 2. This gives a deviation because any player from front 2 could move to compete in front 1 and strictly reduce their cost (they now win front 1 and still win front 2). Similar reasoning holds for the second case.
[TABLE]
Here, type wins by at least one on front 1 and wins by exactly one on front 2. This gives a deviation because type is losing in front 1, and can move to front 2 where it will tie (and still lose front 1). Similar reasoning holds for the second case.
The full proof concludes by proving that any within the median-critical region must satisfy one of the cases, meaning that the arrangement must be unstable. ∎
Finally, we address the question of pairs with (so that they are not addressed by Lemma 2), but which also do not fall in the median-critical region (so they are not addressed by Lemma 3). Lemma 4 examines this case and constructively shows that it is always possible to find a stable arrangement of agents onto fronts, assuming fronts have equal weights.
Lemma 4**.**
Any other number of agents () with (besides those in the median-critical region) always has a stable arrangement, given equal weights and cost satisfying Lemma 1.
Proof sketch.
We will show constructively that it is possible to create an arrangement satisfying the following criteria:
- •
For every front where there is more than 1 agent, type and type tie exactly.
- •
Every other front has exactly one agent, which can be either type or type .
This type of construction is stable by the following reasoning: None of the single agents can move (they can’t leave an front empty). Additionally, no agent on an front with multiple agents wishes to leave - they would go from winning a single front and losing another, to losing on that front and tying on another, which gives equal costs when weights are equal.
Informally, we will describe how this algorithm works. If , then we place exactly one agent on each front, which satisfies the construction criteria.
If , and is odd, then the algorithm places equal numbers of players on front 1, and places exactly one player on every remaining front. On the other hand, if is even, then the algorithm places equal numbers of players on fronts 1, places exactly 1 of each player type on item 2, and then places exactly one player on every remaining front. The distinction between whether is even or odd is necessary to ensure that the total number of players placed sums up to . Algorithm 2 describes this algorithm formally. ∎
Lemma 5, below, demonstrates why the equal weights condition in Lemma 4 is necessary. If even the weight of one front is even slightly higher than the weight in another, there exist conditions where no stable arrangement exists.
Lemma 5**.**
Set , with and , and and cost satisfying Lemma 1. Then, the arrangement proposed by Lemma 4 is not stable, and moreover, there is no possible stable arrangement.
Proof sketch.
In this proof, we will use the notation
[TABLE]
to illustrate that there are 4 fronts in total, two of which with exactly 2 players on it (one of each type), and two fronts with exactly one player (one front with a single type player, and one front with a single type player). Lemma 4 would suggest that the arrangement in Equation 1 would be stable. If , then this would be satisfied: none of the singleton players could move, and none of the players wish to move. If a player of type from front 2 moved to compete in front 1, they would go from experiencing cost to experiencing cost: which is identical when . However, when , then this move does reduce cost, meaning that the original arrangement was unstable. The full proof, in Appendix A, completes the proof by showing that no other possible arrangement is stable. ∎
Finally, Lemma 6 extends Lemma 4 by relaxing the requirement that the front weights be exactly equal. Instead, this proof shows that it is sufficient to have the two fronts with highest weight be equal, while no other two fronts differ in weight by more than a factor of 2. Taken together, these lemmas characterize the stability of the Private Blotto game with median outcome function and more agents than fronts.
Lemma 6**.**
[Extension of Lemma 4] Any other number of agents () with (besides those in the median-critical region) always has a stable arrangement, given cost satisfying Lemma 1 and weights in descending order satisfying .
4.2 Mean outcome
Next, in this section we will explore the setting where again there are more agents than fronts (), but where instead the mean outcome function is used. In Section 4.1 we were able to completely (constructively) classify the stable arrangements for median outcome. By contrast, in this section we will show that the pattern of stable arrangements is much more chaotic.
First, we explore the most direct question of when a stable arrangement exists. Even this question turns out to be surprisingly challenging. Figure 1 numerically explores when a stable arrangement exists for fronts222Code to reproduce figures and numerical examples is available at https://github.com/kpdonahue/private_blotto. . The axes represent the total number of type and type agents that are present, with a red dot appearing at point if no possible stable arrangement involving that number of agents exists. The plots in different panels of the figure vary in the weights given to each of the items. Note that the patterns across each of the plots seem to vary wildly. For equal weights (), there appears to be a structured pattern in when a stable arrangement exists and when one does not. As the weights become imbalanced, this pattern immediately disintegrates, resulting in a series of chaotic clusters.
However, the unpredictable patterns in Figure 1 actually obscures a deeper symmetry within the model. As Theorem 1 below shows, if players were allowed to be allocated fractionally over fronts, then the stable arrangement would always be exactly equal to proportional allocation over fronts (according to each front’s weight):
Theorem 1**.**
For fronts with two types of agents, and with mean outcome and satisfying the conditions of Lemma 1, if players are allowed to be allocated fractionally over fronts, then the stable arrangement is always given by .
Note that, in general, Theorem 1 does not imply that stable arrangement for the standard (integer-valued) Private Blotto games will be close to proportional. While Theorem 1 can be extended to show that the fractional Private Blotto game convex, it is known that in general, the minimum of a convex function, when restricted to integer values may be arbitrarily far from the minimum of the same convex function over real numbers. However, it turns out that, for the Private Blotto game with equal weights, it is true that integer-valued stable arrangements are “close” to proportional. This idea is formalized in Theorem 2:
Theorem 2**.**
Assume fronts with equal weights and two types of agents, and with mean outcome and satisfying the conditions of Lemma 1. Then, any arrangement that is stable must be “close” to proportional: for .
Proof sketch.
In order to prove this result, we will show that any arrangement that is not “close” to proportional must have at least one agent that wishes to move. An arrangement is “close” to proportional if:
[TABLE]
At a high level, our proof technique involves finding a “large” front (with more agents) and a “small” front (with fewer agents). We will then pick whichever agent type is more represented in the larger front, and move exactly one agent to the smallest front. We will then give three sets of sufficient conditions so that this strictly reduces cost for the agent type that moves. We will then show that if none of the three sufficient conditions are met, all fronts must be “close” to proportional.
Selecting the fronts: First, we will consider all pairs of fronts such that . One feature of this pair that we will consider is the gap in type prevalence, which is given by:
[TABLE]
Note that if one type has positive gap in type prevalence, then the other has negative gap in type prevalence, because:
[TABLE]
WLOG, we will assume that type has positive or 0 gap in type prevalence. If the gap in type prevalence is 0 (both fronts have exactly equal proportions of player types), then we will again WLOG assume that type makes up a larger share of players, or . Given this assumption, we will show that players of type could always reduce its cost by moving a single agent from front to front (unless all fronts are close to proportional).
Costs: Type ’s cost associated with the fronts and is:
[TABLE]
Its cost associated with these fronts after moving a single agent from is given by:
[TABLE]
which implies that its cost decreases with this move whenever:
[TABLE]
Or:
[TABLE]
Equation 2 is the central condition we will be studying in this proof. First, we will present several sufficient conditions for when Equation 2 is satisfied, so a player wishes to move. Then, we will show that whenever none of those sufficient conditions are satisfied, all fronts must be “close” to proportional.
Sufficient condition 1: Fronts differ by at least 2, positive gap in type prevalence There are a few conditions where we can immediately see that Equation 2 is satisfied. First, we divide this equation into two separate components. The first is given by:
[TABLE]
By prior reasoning, this is satisfied by the assumption that as given by type having positive gap in type prevalence.
Next, we will consider the second component of Equation 2:
[TABLE]
Note that this is not required by how we selected fronts (all we require is ). However, in the event that both Equation 4 holds, and either Equation 3 or Equation 4 is satisfied strictly, then Equation 2 holds strictly and type players have an incentive to move from front 1 to front 2.
In the full proof, we consider other sufficient conditions where Equation 4 and 3 are not both strictly satisfied, and yet a type player still wishes to move from front to . Finally, we conclude by showing that any arrangement that fails to satisfy one of the sufficient conditions must have all fronts having arrangements that are “close” to proportional. ∎
Theorem 2 suggests the following hypothesis, which is that the integer-valued Private Blotto game with arbitrary weights over fronts still results in arrangements that are “close” to proportional.
Hypothesis 1**.**
Any stable arrangement in the integer Private Blotto game with mean outcome function must be “close” to proportional: for .
From numerical simulations Hypothesis 1 appears to hold for our functional form; we have not found counter-examples computationally, and we conjecture that it always holds.
4.3 Misallocated effort
Finally, in this section we will compare the stable arrangements given either median or mean outcome functions. In particular, we will consider the question of how “bad” stable arrangements might be. This question has been formalized in a variety of ways in previous papers, including Price of Anarchy or Price of Stability [Koutsoupias and Papadimitriou(2009), Anshelevich et al.(2008)]. For example, in a congestion routing game, Price of Anarchy would measure the total congestion for all players in the worst-case stable arrangement, as compared to the arrangement that minimizes total congestion.
However, Private Blotto is modeling a fundamentally different game. In the political debate scenario, for example, Private Blotto is modeling different political actors competing over topics for which they have truly different viewpoints. In this case, society doesn’t necessarily have a concept of which outcome is “right”. However, we might have normative preferences over how we wish these debates to unfold. One concrete example might be something like “political topics should be debated in rough proportion to how important they are to all actors”. For example, if a legislative body spends a week debating the naming of a bridge and only a morning debating the contents of a vital spending bill, we might view that allocation of effort as wasteful. This intuition of “misallocated effort” is formalized below:
Definition 5**.**
Given an arrangement of agents onto fronts, we say it has “misallocated effort” given by the amount of agents that is above or below allocation equal to the weights. That is, misallocated effort is given by:
[TABLE]
One question we will explore is the maximum possible misallocated effort, among any stable arrangements. The results of Lemma 1 require that each front has at least one agent, which Lemma 7 uses to give an immediate upper bound to misallocated effort.
Lemma 7**.**
Given satisfying Lemma 1, each front to have at least one agent, which means that (regardless of the outcome function used), misallocated effort is upper bounded by .
Next, we will obtain tighter bounds for mean and median outcome functions, respectively. First, Lemma 8 shows that a much tighter bound on misallocated effort is possible for mean outcome functions.
Lemma 8**.**
For mean outcomes, if Hypothesis 1 holds then misallocated effort is upper bounded by .
Proof.
This is a direct consequence of Hypothesis 1:
[TABLE]
∎
Next, Lemma 9 shows that worst-case misallocated effort can be much higher for the median outcome function, especially in the case where there are many more agents than fronts. Lemma 7 tells us that misallocated effort is likely to be highest for large . Lemma 9 looks at this specific case and shows that misallocated effort could be as high as , which can be much greater than the bound given in Lemma 8.
Lemma 9**.**
For median outcomes, worst-case misallocated effort is lower-bounded by , given .
Proof.
We rely on the stable arrangements found in the proof of Lemma 4. For , the arrangement starts by placing 2 of each of type players on each front, up until either we run out of type players or reach the st front. We then place the type on the remaining fronts. In the case that , this implies that a single front has type agents. The preconditions of Lemma 4 assume equal weights , which means that misallocated effort is lower bounded by Because we have required and , this bound is at least: . ∎
Overall, the results of this analysis imply that mean outcome functions, rather than median ones, give sharper guarantees that any stable arrangement that exists will result in agents roughly arranging themselves across fronts in proportion to their overall value.
5 Fewer agents than fronts:
In Section 4, we examined the case with more agents than fronts: . In this section, we will explore the other possibility, with fewer agents than fronts. A motivating example for this scenario could be voluntary crowdsourcing on a social media platform, such as Twitter’s BirdWatch [Wojcik et al.(2022)]. [Allen et al.(2022)][Pröllochs(2022)]. Here, there might be many more relevant items (e.g. news articles or tweets) than there are active labelers. Our goal here is to model the set of stable arrangements, again comparing median and mean outcome functions.
In this section, we will drop Assumption 1, as all of our results extend easily to arbitrary numbers of agent types . Because , some fronts will inevitably need to be left empty. Because of this, we will drop the lower bound in Lemma 1 and allow the cost for leaving a front empty to be set arbitrarily. Since Lemma 1 no longer applies, in this section we will see that agent biases are relevant for agents’ strategies.
At a high level, Section 4 showed a wide divergence in behavior between median and mean outcome functions. In this section, we will show that the setting of gives much more similar results between the two outcome functions, though with some differences. The intuition is that both median and mean outcome functions behave identically for fronts with only 1 or 2 agents present. Because there are relatively few agents, compared to the number of fronts, most arrangements will have 1 or 2 agents per front, unless is very small or differences in biases between types is very large.
First, we analyze the most natural arrangement: where there is exactly one agent on each of the most valuable fronts, with the remaining ones left empty. Lemma 10 gives conditions where this type of arrangement is stable, showing that these conditions are identical for both median and mean outcome functions.
Lemma 10**.**
Assume that . WLOG, relabel the fronts in descending order of weight, so . Then, arrange each agent on each front in descending order of : that is, absolute value of bias, so that . Define to be the first element that has an opposite sign from . (If they are all the same sign, then set ). Then, place exactly one agent on each front in descending order of bias magnitude and weight. This arrangement is stable (for either median or mean outcome functions) so long as:
[TABLE]
Next, Lemma 11 studies another natural scenario: where there are exactly two players and at least two fronts. Here, the lemma shows that for both median and mean outcome functions, players either both prefer to compete in the same front, or else both prefer to compete in separate fronts.
Lemma 11**.**
For , with equal weights and either median or mean outcome functions, the two players either both prefer competing on the same front or both prefer each competing on separate fronts. Because their preferences are perfectly aligned, this implies that one of the two arrangements is always stable.
Proof.
Consider two players with bias . The player of type get lower cost when competing in separate fronts whenever:
[TABLE]
On the other hand, the player of type gets lower cost when competing in separate fronts whenever:
[TABLE]
These terms are exactly equivalent, proving the result. ∎
5.1 Median outcome
The previous section described specific conditions for when a Nash stable arrangement exists. Next, we will explore the complementary condition by showing when (for median outcome functions), no stable arrangement exists. Lemma 12 gives a constructive example of settings where any possible arrangement of agents onto fronts is unstable.
Lemma 12**.**
For any with such that , with median outcome, there exists biases and costs such that no NE exists.
Proof sketch.
Set parameters as follows:
- •
fronts, all with equal weight.
- •
players, with given biases: 1 with bias , and with bias .
- •
Cost of 0.3 for leaving an front empty.
The full proof considers any possible arrangement of these players, at least one can reduce its total cost by moving. ∎
5.2 Mean outcome
Next, we will consider the case with mean outcome function. Lemma 13 follows a similar approach to Lemma 12, showing constructively that there exist parameters where no stable arrangement exists. However, as compared to Lemma 12, it has the added requirement of .
Lemma 13**.**
For any such that , with mean outcome, there exists parameters such that no NE exists.
Proof sketch.
The parameters are set identically to in Lemma 12, except with a cost of for leaving an front empty. The proof follows similarly to that for Lemma 12, exploring multiple possible arrangements and showing that none can be stable. ∎
The mean outcome function case explored in Lemma 13 is not exactly analogous to the median case in Lemma 12: there is a gap at players. Lemma 14 complete this gap by showing that the gap in Lemma 13 is inevitable: any possible set of agents must have a stable arrangement, given mean outcomes and equal weights. Note that the case with equal weights is the most interesting one: if weights are allowed to differ, then a stable arrangement can easily be found by setting one front to have extremely high weight, resulting in all agents competing in that front.
Lemma 14**.**
For with mean outcome and equal weights, there is always a stable point.
Proof sketch.
There are five possible arrangements of 3 agents:
[TABLE]
We will use the notation
[TABLE]
to mean that a player of type gets strictly lower cost when it leaves a front it is competing in with type to compete in an empty front. We will say when the player of type does not get strictly lower cost in and say when the player of type gets strictly lower cost in . Often, to be concise, we will drop terms and simply write as the second front is left empty on the lefthand side.
In this proof, we will use the results of Lemma 11 to show that for any pair of agents, they either would both prefer to be competing in the same front or else both prefer to be competing in different fronts. Next, we will show that if any two pairs of agents prefer competing in the same front (as opposed to competing in different fronts), then it is the two agents with most dissimilar biases (e.g. given ). In order to show this, we will write out the cost that an agent of type gets for every possible arrangement. Because agents are interchangeable, this gives costs for every other agent in different arrangements, up to relabeling.
[TABLE]
The second and third lines in the equations above show us that
[TABLE]
exactly whenever:
[TABLE]
[TABLE]
which is whenever is further from than is. In this analysis, WLOG we will say that are the most dissimilar (because agents are interchangeable, this is true up to relabeling). Note that this implies that and can never be stable, because both would prefer being together to being with .
Next, we can analyze some cases:
Case 1: If the , then is stable.
We know from our prior reasoning that if the two most dissimilar players do not wish to compete in the same front, then no other pair of players do. This means that the arrangement with one agent per front is stable.
Case 2: If:
[TABLE]
then is stable. This is because the pair doesn’t wish to split up and the type player doesn’t wish to join.
Case 3: If no player wishes to leave to compete in an empty front, then is stable.
This is true simply by the statement: if no player wishes to move, then this arrangement must be stable.
Case 4: In this case, we will assume that:
[TABLE]
We will also assume that at least one of wishes to leave . WLOG, we will assume it is (up to relabeling). Note that this implies none of these arrangements can be stable:
[TABLE]
Additionally, by our prior reasoning, if players prefer competing together to separate, then we know that no other pair can be stable:
[TABLE]
The cycle that is given by this set of patterns is given by:
[TABLE]
We will show that this type of cycle cannot exist.
Note that the first move occurs when:
[TABLE]
The second move occurs when:
[TABLE]
And the third move occurs when:
[TABLE]
Putting these together, this implies:
[TABLE]
[TABLE]
In the full proof, we complete our analysis by showing that it is impossible to set so as to satisfy both of these inequalities simultaneously. ∎
6 Discussion
In this paper, we have proposed and analyzed the Private Blotto game, a multi-player game involving competition over fronts with different values. We focused on the impact of the outcome function on whether Nash stable arrangements exist. For the case with more agents than fronts, in Section 4 we showed the existence of a median-critical set of parameters such that no stable arrangement is possible within this set. We additionally showed constructively that a stable arrangement is always possible for any set of parameters outside of the median-critical set (assuming weights on fronts that are equal or bounded in difference). However, stable arrangements may involve high degrees of “misallocated effort” where agents are distributed across fronts in ways that are far proportional to the front’s value. For mean outcome rules, we showed that any stable arrangement must involve bounded misallocated effort, but patterns in which stable arrangement exist and fail to exist do not have a clean theoretical characterization. In Section 5 we analyzed the complementary case with fewer agents than fronts. In this case, median and mean outcome functions have much more similar results. For both, we constructively show conditions for the existence of a stable solution, as well as showing that there exist parameters such that no stable arrangement can exist.
While in this work we addressed the primary question of Nash stability, there are multiple interesting extensions into the Private Blotto game. One natural extension could be to explore a combined Colonel Blotto and Private Blotto game, with some of the “types” being centrally-coordinated Colonel Blotto with others being distributed Private Blotto. The Private Blotto players would be strictly weaker than the Colonel Blotto ones, but might be able to perform competitively given increased numbers of agents. Another generalization of Private Blotto could allow agents to coordinate with up to other agents. For example, for , an arrangement would fail to be Nash stable if two agents, working together, could move and improve both of their utility. This modification would only make Nash stability harder to achieve, but could be interesting to analyze. Finally, some Colonel Blotto games allow for players to differ in their subjective valuations or weights for each front - if, for example, certain issues are more valuable to certain political groups than to others [Schwartz et al.(2014b)]. Extending this model could allow for interesting variations in patterns of Nash stability.
Acknowledgements
This work was supported in part by a Simons Investigator Award, a Vannevar Bush Faculty Fellowship, MURI grant W911NF-19-0217, AFOSR grant FA9550-19-1-0183, ARO grant W911NF19-1-0057, a Simons Collaboration grant, a grant from the MacArthur Foundation, and NSF grant DGE-1650441. We are extremely grateful to Maria Antoniak, Sarah Dean, Jason Gaitonde, and the AI, Policy, and Practice working group at Cornell for invaluable discussions.
Appendix A Proofs
See 1
Proof.
In other words, we want to ensure that for any player , the cost of competing in any front (leaving any other front empty) is higher than the cost of leaving front to competing in front alone:
[TABLE]
First, we will analyze the case with mean outcome function. For an agent of type , the cost it experiences from a front with agents of type and agents of type is given by:
[TABLE]
By identical reasoning, the cost to an agent of type is:
[TABLE]
Note that this construction immediately tells us that agent strategy must be independent of biases. For every front with , an agent’s cost is solely a function of and , scaled by a constant factor of .
Next, we will work on determining so that no front will ever be left empty. Again, we wish to show that:
[TABLE]
The worst-case scenario (where this inequality is hardest to satisfy) occurs when the empty front has very low weight, relative to the front it is competing in (). If we consider a reference player of type , with players of type on front and of type , then this becomes:
[TABLE]
[TABLE]
[TABLE]
Next, we’ll upper bound the term on the RHS. The RHS shrinks with , so we can lower bound this by setting . We know that because we have assumed there is at least one player of type that wishes to move from the given front. The condition simplifies to:
[TABLE]
We similarly must have (or else we’re just modeling a single player of type move from one front to another). If we set , then this goes to 1/2, which gives the desired condition. Intuitively, this tells us that we need that the cost of leaving something unlabeled is greater than half the distance between the two biases.
Next, we will consider the case where the outcome function is equal to the median. Again, we wish to show that:
[TABLE]
We will analyze multiple different cases for the potential outcome functions and . Again, we will look from the perspective of a type agent on front considering moving to another front that is empty:
- •
and . The inequality becomes:
[TABLE]
which is satisfied automatically.
- •
and . The inequality becomes:
[TABLE]
[TABLE]
- •
and . The inequality becomes:
[TABLE]
[TABLE]
- •
and . The inequality becomes:
[TABLE]
which is always satisfied.
The only inequality that isn’t automatically satisfied is , which is the same inequality as for the mean outcome function, and satisfied by the same reasoning.
Finally, we will show that agents’ incentives are independent of biases . For any agent with bias and any set , the outcome function has three possible values: (giving cost to agent of 0), in the event of ties (giving cost to agent of ), or (giving cost to agent of ). All of these are simply scaled values of , which means incentives are independent of the values . ∎
See 3
Proof.
Note that if we have , then the last criteria gives us , which isn’t achievable.
To start out with, we will describe a few arrangements where a deviation is always possible. For notation, we will use to denote an arrangement where there are at least as many type players as type on front 1, for example. We will only refer to fronts 1 and 2 for convenience, but these results hold for any labeled fronts.
[TABLE]
This gives an arrangement where type loses by 1 on front 1 and wins by at least 2 on front 2. This gives a deviation because any player from front 2 could move to label the front 1 and strictly reduce their cost (they now tie in the first and still win in the second). Similar reasoning holds for the other case: the type player from front 2 could move to label front 1 and strictly reduce their cost.
[TABLE]
Here, the players tie on front 1 and type wins by at least 2 on front 2. This gives a deviation because any player from front 2 could move to label front 1 and strictly reduce their cost (they now win front 1 and still win front 2). Similar reasoning holds for the second case.
[TABLE]
Here, type wins by at least one on front 1 and wins by exactly one on front 2. This gives a deviation because type is losing in front 1, and can move to front 2 where it will tie (and still lose front 1). Similar reasoning holds for the second case.
Next, we will show that if we have any arrangement satisfying the preconditions (), then at least one of these cases must occur, meaning that the arrangement must be unstable.
First, let’s suppose that we have at least one tie somewhere: . By Case 2, we know that we can’t have any player types win by 2 elsewhere (or else one of the agents could move and win at front 1, while still winning elsewhere). That means that we need:
[TABLE]
We will show that satisfying the second condition is impossible. First, summing over all of the fronts gives:
[TABLE]
[TABLE]
Recall that (there’s a tie), so this reduces to:
[TABLE]
[TABLE]
However, we also have , so this becomes:
[TABLE]
[TABLE]
which is a contradiction. This implies that no arrangement where players tie can be stable (and satisfy the preconditions).
Next, let’s consider the case where we have (type loses front 1 by exactly 1). Case 1 tells us that we cannot have elsewhere (cannot have that type wins by at least two elsewhere), so we must have . Again, we can sum:
[TABLE]
[TABLE]
Substituting in for gives us:
[TABLE]
[TABLE]
[TABLE]
which again contradicts .
Finally, let’s consider where we have or . By Case 1, we must have for all . However, we will show that this implies a contradiction with other stability analayses.
First, let’s consider the case where for all , so type wins every front by one except the first one, which it loses by exactly one. We will show that this violates the preconditions of this lemma:
[TABLE]
[TABLE]
[TABLE]
[TABLE]
[TABLE]
[TABLE]
If we combine this with our condition, we get:
[TABLE]
[TABLE]
[TABLE]
which cannot be satisfied because we need at least one front.
Next, we’ll consider the case where for at least one front. From our previous analysis of Case 2, we know that we cannot have an exact tie ().
This implies that we must have at least one front such that , along with front 1 which has . Equivalently, we can write this as . However, this exactly implies the condition in Case 3, which is also unstable. ∎
See 4
Proof.
We will show constructively that it is possible to create an arrangement satisfying the following criteria:
- •
For every front where there is more than 1 agent, type and type tie exactly.
- •
Every other front has exactly one agent, which can be either type or type .
This type of construction is stable by the following reasoning:
- •
None of the single agents can move (they can’t leave an front empty).
- •
No agent on an front with multiple agents wishes to leave - they would go from winning a single front and losing another, to losing on that front and tying on another, which gives equal costs when weights are equal.
Algorithm 2 considers this case. Informally, we will describe how it works.
If , then we place at most one agent on each front, which satisfies the construction criteria.
Next, we consider the case where is even. We calculate and place of type and type players each on front 0. By the assumptions of this lemma, we know that , which means that . Then, we place exactly 1 of type and type on front 1. For every other front, have with exactly one type agent, exactly with exactly one type agent. The total number of fronts:
[TABLE]
as desired.
Next, if is odd, we know that is even. We then calculate . Then, set (we address the case where at the end of this proof). Every other front has exactly one agent, with of them having type , and having type . Note that the total number of agents adds up to the right amount and that (by construction) every front is labeled. 1 front has many players, have type only, have type only, and together this sums to:
[TABLE]
Finally, we consider the case where , which means . By assumptions of this lemma, we know that . Taken together, this tells us that:
[TABLE]
This can only be satisfied by setting . However, this means that:
[TABLE]
However, this conflicts with the assumption that is odd, which means this situation can never occur.
∎
See 5
Proof.
In this proof, we will use the notation
[TABLE]
to illustrate that there are 4 fronts in total, two of which with exactly 2 players on it (one of each type), and two fronts with exactly one player (one front with a single type player, and one front with a single type player). Lemma 4 would suggest that the arrangement
[TABLE]
would be stable. If , then this would be satisfied: none of the singleton players could move, and none of the players wish to move. If a player of type from front 2 moved to label front 1, they would go from experiencing cost
[TABLE]
to experiencing cost:
[TABLE]
which is identical when . However, when , then this move does reduce cost, meaning that the original arrangement was unstable.
We will further show that no possible arrangement is stable. Note that fronts have identical weight and are therefore interchangable. Because and , if no front can be left empty, then no front can have more than 3 players. We will consider each case based on the maximum number of agents on a front.
- Case 1
No more than 3 agents on one front: The arrangement goes to because type ’s cost goes from to . (Type continues to win on item 2 and now also ties on item 3). Identical reasoning holds for any symmetrical case with 3 players of the same type on any item.
- Case 2
No more than 3 agents on one front: The arrangement moves to : the type player on front 1 can move to compete in front 3, which takes its cost from to . (Type continues to lose item 1 and now ties on item 3). Identical reasoning holds for the symmetric case with type , as well as for arrangements where player has 3 players: for example, the arrangement goes to because the type players cost goes from to .
- Case 3
No more than 2 agents on a front: Given the arrangement , as described above, either type or type players from front 1 would wish to move to compete in front 1. This is the case analyzed at the beginning of the proof, where we showed this is unstable for .
- Case 4
No more than 2 agents on a front: The arrangement goes to because player type ’s cost goes from to . Similarly, any arrangement with exactly 2 agent of the same type on an front must leave at least one other front with exactly 1 agent of the opposite type, and will be unstable for the same reasons.
This description is exhaustive: If there is a maximum of 3 agents on a front, then they must all be of the same type (Case 1) or two of the same type, and one of another type (Case 2). If there is a maximum of 2 agents on a front, then if one front has exactly one agent of each type, then another front must have exactly one agent of each type (Case 3) or must have exactly 2 agents of one type (Case 4). ∎
See 6
Proof.
This proof is very similar to that of Lemma 4, so we will simply note key differences in the analysis. Throughout, we will assume that for (the fronts are organized in descending order of weight).
First, we will suppose that either:
[TABLE]
Again, we will allocate agents according to Algorithm 1. This arrangement is stable by identical reasoning: for each front, each type wins by at least 2, so no single agent acting alone can change the outcome, regardless of weights.
- •
For every item where there is more than 1 labeler, type and type tie exactly.
- •
Every other item has exactly one labeler, which can be either type or type .
- •
Any item where players tie has higher or equal weight to any item where a single player wins.
Next, we consider the case where . We will show constructively that it is possible to create an arrangement satisfying the following criteria:
- •
For every front where there is more than 1 agent, type and type tie exactly.
- •
Every other front has exactly one agent, which can be either type or type .
- •
Any front where players tie has higher or equal weight to any front where a single player wins.
This type of construction is stable by the following reasoning:
- •
None of the single agents can move (they can’t leave an front empty).
- •
No agent on an front with multiple agents wishes to leave - they would go from winning a single front and losing another, to losing on that front and tying on another, which gives equal costs when weights are equal.
- •
No agent on a front with multiple agents wishes to leave: they would go from tying on front and losing on front to losing on front and tying on front . This gives higher or equal cost when , which is satisfied by construction.
Note that we require by the example given in Lemma 5. ∎
See 1
Proof.
In order to show this, we will look at a relaxed (continuous) version of this problem. In this relaxed version of the problem, we will assume that, instead of agents coming in integer units, they can be allocated fractionally across items.
The cost to player of type is given by . The partial derivative of this with respect to is:
[TABLE]
In order for us to be at a stable point, we need that the derivative wrt must be equal to the derivative wrt for any and (identical criteria for ). If the first is not satisfied, then type could strictly reduce its cost by changing its allocations between fronts, and if the second is not satisfied, then type could again reduce its cost by allocating more agents onto fronts. We can enforce both of these by setting:
[TABLE]
[TABLE]
By identical reasoning, this means that for player , the derivative of its cost wrt is:
[TABLE]
Setting these equal to each other gives:
[TABLE]
[TABLE]
Substituting in for the overall derivative gives us that:
[TABLE]
Cancelling common factors on each side of the equality gives:
[TABLE]
Or . We can apply the equality to obtain:
[TABLE]
which implies and symmetrically. This tells us that the only possible equilibrium is when both and players have set exactly proportional to the weight each front has. ∎
See 2
Proof.
In order to prove this result, we will show that any arrangement that is not “close” to proportional must have at least one agent that wishes to move. An arrangement is “close” to proportion if:
[TABLE]
For any arrangement falling outside of these parameters, we will give a move that reduces cost for at least one player type. At a high level, this will involve finding a “large” front (with more agents) and a “small” front (with fewer agents). We will then pick whichever agent type is more represented in the larger front, and move exactly one agent to the smallest front. We will show that, almost always, this strictly reduces cost for the agent type that moves. In the cases where such a move would not strictly reduce cost, we show that all fronts must be “close” to proportional.
First, we will consider all pairs of fronts such that . One feature of this pair that we will consider is the gap in type prevalence, which is given by:
[TABLE]
Note that if one type has positive gap in type prevalence, then the other has negative gap in type prevalence, because:
[TABLE]
WLOG, we will assume that type has positive or 0 gap in type prevalence. If the gap in type prevalence is 0 (both fronts have exactly equal proportions of player types), then we will again WLOG assume that type makes up a larger share of players, or . Given this assumption, we will show that players of type could always reduce its cost by moving a single agent from front to front (unless all fronts are close to proportional).
Costs: Type ’s cost is:
[TABLE]
Its cost after moving a single agent from is given by:
[TABLE]
So, its cost decreases whenever:
[TABLE]
Or:
[TABLE]
Equation 2 is the central condition we will be studying in this proof. First, we will present several sufficient conditions for when Equation 2 is satisfied, so a player wishes to move. Then, we will show that whenever none of those sufficient conditions are satisfied, all fronts must be “close” to proportional.
Sufficient condition 1: Fronts differ by at least 2, positive gap in type prevalence There are a few conditions where we can immediately see that Equation 2 is satisfied. First, we divide this equation into two separate components. The first is given by:
[TABLE]
[TABLE]
By prior reasoning, this is satisfied by the assumption that as given by type having positive gap in type prevalence.
Next, we will consider the second component of Equation 2:
[TABLE]
which is satisfied exactly whenever:
[TABLE]
Note that this is not required by how we selected fronts (all we require is ). However, in the event that both Equation 4 holds, and either Equation 3 or Equation 4 is satisfied strictly, then Equation 2 hold strictly and type players have an incentive to move from front 1 to front 2.
Next, we will consider other cases where Equation 4 and 3 are not both strictly satisfied, and yet a type player still wishes to move from front to .
Sufficient condition 2: Larger and smaller front have same number of players, First, we will consider the case where (the two fronts have equal numbers of players). Equation 2 simplifies to
[TABLE]
We will show that so long as , then this inequality must hold. Note that this condition implies . Substituting in for and tells us that we wish to show:
[TABLE]
[TABLE]
[TABLE]
[TABLE]
[TABLE]
Because , we know that , as desired.
When this sufficient condition is not met, we have that . Because we assume that type has positive gap in type prevalence, this means that and in order to ensure .
Sufficient condition 3: Larger front has exactly 1 more agent than smaller, Next, we will consider the case where the larger front has exactly 1 more player than the smaller front, or . Then, the condition for a player of type A wanting to move (Equation 2) becomes:
[TABLE]
which simplifies down to:
[TABLE]
If we have , then we must have:
[TABLE]
which implies that:
[TABLE]
as desired.
When this sufficient condition is not met, we have that . Because we assume that type has positive gap in type prevalence, this means that and in order to ensure .
**In absence of sufficient conditions, all fronts are “close” to proportional
**Finally, we will consider the case where none of the sufficient conditions hold. By taking the negative of previous cases, we know that:
By sufficient condition 1, there is no pair of fronts with . 2. 2.
By sufficient condition 1 for any pair of fronts with , we must have zero gap in type prevalence. 3. 3.
By sufficient condition 2, for any pair of fronts with , we must have . 4. 4.
By sufficient condition 3, for any pair of fronts with , we must have .
Each of these cases (except for item 2) results in an arrangement where for any pair of fronts , we have
[TABLE]
If we are also able to show that this holds for item 2 (when ), then we will know that any pair of fronts differs in by at most 1. This would mean that every front has either or and or . The average number of players per fronts must lie somewhere between , which means that every front is “close” to proportional.
We will conclude our proof by showing that examining item 2 and showing that we must also have .
In this case, we assume that . Note that by our analysis of Case 1, a type player wishes to move automatically if Equation 3 holds strictly. In this case, we will assume that it does not hold strictly and derive additional conditions on . Equation 3 becomes an equality:
[TABLE]
This implies that fronts have exactly the same proportion of type and type players (a zero gap in type prevalence). This means that we can write . In how we selected fronts, we assumed WLOG that in the case that the gap in type prevalence is 0, players of type make up a greater share of players on both fronts. This implies that . We can then write:
[TABLE]
Substituting in for gives:
[TABLE]
[TABLE]
The only value for which this results in an integer is , which implies , and , or as desired. ∎
See 10
Proof.
For simplicity, we will say that front is labeled by player with bias . In order for this arrangement to be stable, we need to know that no player wishes to label a different front .
First, we can immediately see that no player would wish to label an front that is currently empty. By the way we have arranged players, we know that for , so labeling that front would have less impact on overall cost.
Next, we need to show that no player wishes to label a different front that already has a label on it.
No player wishes to move to another front that already has a label so long as so long as:
[TABLE]
[TABLE]
[TABLE]
[TABLE]
The RHS is lower bounded by setting . The LHS is upper bounded by plugging in for (the largest magnitude) and (the next largest magnitude with an opposite sign), meaning that the LHS will be equal to rather than ). This gives us:
[TABLE]
∎
See 12
Proof.
Set parameters as follows:
- •
fronts, all with equal weight.
- •
players, with given biases: 1 with bias , and with bias .
- •
Cost of 0.3 for leaving an front empty.
First, we will note that any arrangement where 3 or more players are labeling a single front (front ) must fail to be stable. Because , we know that every arrangement must leave at least one front empty, for a cost of 0.3.
- •
First, suppose that the front in question has only players with bias labeling it. Then, any player can leave front and instead label an empty front . The median of front remains the same and the median of becomes (eliminating empty cost), which reduces the cost for players with bias .
- •
Next, consider the other case where the front in question is labeled by the player with bias 1. We know that there are at least 2 players with bias , since there are at least 3 players total. If the player with bias 1 leaves and instead labels , then the median remains -0.5. However, again, this now eliminates the penalty for leaving empty.
Now, we consider all arrangements where there are no more than 2 players labeling an front. We similarly know that if there are 2 players of bias labeling an front, then they can reduce their cost by having one leave to label an empty front (maintaining the same median and eliminating empty cost). Therefore, we can focus on how the player with bias is arranged. We will find that considering only the first 3 fronts will be sufficient. For conciseness, we will use the notation to mean that three players are all labeling a single front each (one player with bias 1, two players with bias ).
- •
goes to . Originally, the -0.5 players have cost . The player wishes to move, which produces medians , which gives it cost become .
- •
goes to . The player wishes to move again. Before, the medians are , giving it a cost of . After it moves, the medians are , which gives it a cost of .
- •
goes to . The 1 player wishes to move. Before, the medians are , which gives it a cost of . After it moves, the medians are , which gives it a cost of .
- •
goes to . The player wishes to move. Before, the medians are , which gives it cost . After it moves, the medians are , which gives it cost .
∎
See 13
Proof.
Set parameters as follows:
- •
fronts, all with equal weight.
- •
players, with given biases: 1 with bias , and with bias .
- •
Cost of for leaving an front empty.
Because , we know that every arrangement must leave at least one front empty, for a cost of . First, we will consider the case where at least one front has 3 or more agents on it.
- •
First, suppose that the front with 3 or more players labeling it has only players with bias labeling it. Then, any player can leave front and instead label an empty front . The mean of front remains the same and the mean of becomes (eliminating empty cost), which reduces the cost for players with bias .
- •
Next, consider the other case where the front in question is labeled by the player with bias 1. Say that there are players with bias -0.5. We know that there since there are at least 3 players total.
- –
Currently, the cost to players of type is:
[TABLE]
If one player of type goes to label the empty front, the cost becomes:
[TABLE]
The type player wishes to move whenever:
[TABLE]
[TABLE]
Note that the lefthand side is decreasing in , which is where the lower bound comes from. Note that if , the lower bound becomes .
- –
Currently, the cost to players of type is:
[TABLE]
If one player of type goes to label the empty front, the cost becomes:
[TABLE]
The type player wishes to move whenever:
[TABLE]
[TABLE]
- –
We are in the scenario where with (the sole type player is on front ). Suppose that there exists another front with . Then, the type player can reduce its cost by moving from front to front whenever:
[TABLE]
[TABLE]
For , the term inside the absolute value is given by:
[TABLE]
which is increasing in , meaning that the inequality is always satisfied.
Now, we consider all arrangements where there are no more than 2 players labeling an front. We similarly know that if there are 2 players of bias labeling an front, then they can reduce their cost by having one leave to label an empty front (maintaining the same mean and eliminating empty cost). Therefore, we can focus on how the player with bias is arranged. We will find that considering only the first 3 fronts will be sufficient. For conciseness, we will use the notation to mean that three players are all labeling a single front each (one player with bias 1, two players with bias ).
- •
goes to . Originally, the -0.5 players have cost . The player wishes to move, which produces means . (The second front has undefined mean, given that it has no labels). This gives the players cost . This is lower if , which is satisfied for .
- •
goes to . The player wishes to move again. Before, the means are , giving it a cost of . After it moves, the means are , which gives it a cost of . This is lower if , or .
- •
goes to . The 1 player wishes to move. Before, the means are , which gives it a cost of . After it moves, the means are , which gives it a cost of , which is always satisfied.
- •
goes to . The player wishes to move. Before, the means are , which gives it cost . After it moves, the medians are , which gives it cost , which is always satisfied.
These cases cover all possible arrangements, showing that there always exists a agent that wishes to move. ∎
See 14
Proof.
There are five possible arrangements of 3 agents:
[TABLE]
We will use the notation
[TABLE]
to mean that a player of type gets strictly lower cost when it leaves a front it is competing in with type to compete in an empty front. We will say when the player of type does not get strictly lower cost in and say when the player of type gets strictly lower cost in . Often, to be concise, we will drop terms and simply write as the second front is left empty on the lefthand side.
First, we claim that:
[TABLE]
Note that the lefthand term of Equation 7 is satisfied whenever:
[TABLE]
[TABLE]
The righthand term of Equation 7 is satisfied whenever:
[TABLE]
[TABLE]
These terms are exactly equivalent, showing Equation 7 is satisfied. We will use this result in the following analysis.
Next, we will show that if any two pairs of agents prefer competing in the same front (as opposed to competing in different fronts), then it is the two agents with most dissimilar biases (e.g. given ). In order to show this, we will write out the cost that an agent of type gets for every possible arrangement. Because agents are interchangable, this gives costs for every other agent in different arrangements, up to relabeling.
[TABLE]
The second and third lines in the equations above show us that
[TABLE]
exactly whenever:
[TABLE]
[TABLE]
which is whenever is further from than is. In this analysis, WLOG we will say that are the most dissimilar (because agents are interchangable, this is true up to relabeling). Note that this implies that and can never be stable, because both would prefer being together to being with .
Next, we can analyze some cases:
**Case 1:
**If the , then is stable.
We know from our prior reasoning that if the two most dissimlar players do not wish to compete in the same front, then no other pair of players do. This means that the arrangement with one agent per front is stable.
**Case 2:
**If:
[TABLE]
then is stable. This is because the pair doesn’t wish to split up and the type player doesn’t wish to join.
**Case 3:
**If no player wishes to leave to compete in an empty front, then is stable.
This is true simply by the statement: if no player wishes to move, then this arrangement must be stable.
**Case 4:
**In this case, we will assume that:
[TABLE]
We will also assume that at least one of wishes to leave . WLOG, we will assume it is (up to relabeling). Note that this implies none of these arrangements can be stable:
[TABLE]
Additionally, by our prior reasoning, if players prefer competing together to separate, then we know that no other pair can be stable:
[TABLE]
The cycle that is given by this set of patterns is given by:
[TABLE]
We will show that this type of cycle cannot exist.
Note that the first move occurs when:
[TABLE]
[TABLE]
[TABLE]
The second move occurs when:
[TABLE]
[TABLE]
[TABLE]
And the third move occurs when:
[TABLE]
[TABLE]
Putting these together, this implies:
[TABLE]
[TABLE]
We will show that these pairs of inequalities cannot be simultaneously satisfied. Set:
[TABLE]
Then, Equation 9 becomes
[TABLE]
[TABLE]
For Equation 8 note that the LHS is given by:
[TABLE]
where the equality is because:
[TABLE]
as desired. The RHS is given by:
[TABLE]
Which is given by:
[TABLE]
So together, Equations 8 and 9 become:
[TABLE]
[TABLE]
Note that if both are positive, Eq. 10 becomes:
[TABLE]
[TABLE]
which directly contradicts Eq. 11 (). Similarly, if both are negative, then Eq. 10 becomes:
[TABLE]
[TABLE]
which contradicts Eq. 11 which has become
[TABLE]
Finally, let’s consider the case where one of is positive and one is negative. WLOG, say . Then, we know that the LHS of Eq. 10 is positive (because so , which makes the overall positive). This means that the RHS of Eq. 10 must be positive. If , then this becomes:
[TABLE]
[TABLE]
which is violated because . If , then this becomes:
[TABLE]
[TABLE]
which again violates , with given by Eq. 11.
Taken together, these cases cover all possible combination of stable relations, showing that there can never be a cycle (and must always have a stable point).
∎
The reference list from the paper itself. Each links out to its DOI / PubMed record.
- 1[1]
- 2[Ahmadinejad et al . (2019)] Amir Mahdi Ahmadinejad, Sina Dehghani, Mohammad Taghi Hajiaghayi, Brendan Lucier, Hamid Mahini, and Saeed Seddighin. 2019. From duels to battlefields: Computing equilibria of Blotto and other games. Mathematics of Operations Research 44, 4 (2019), 1304–1325.
- 3[Allen et al . (2022)] Jennifer Allen, Cameron Martel, and David G Rand. 2022. Birds of a Feather Don’t Fact-Check Each Other: Partisanship and the Evaluation of News in Twitter’s Birdwatch Crowdsourced Fact-Checking Program. Association for Computing Machinery, New York, NY, USA. https://doi.org/10.1145/3491102.3502040 · doi ↗
- 4[Anbarcı et al . (2020)] Nejat Anbarcı, Kutay Cingiz, and Mehmet S Ismail. 2020. Proportional resource allocation in dynamic n-player Blotto games. ar Xiv preprint ar Xiv:2010.05087 (2020).
- 5[Anshelevich et al . (2008)] Elliot Anshelevich, Anirban Dasgupta, Jon Kleinberg, Éva Tardos, Tom Wexler, and Tim Roughgarden. 2008. The price of stability for network design with fair cost allocation. SIAM J. Comput. 38, 4 (2008), 1602–1623.
- 6[Boix-Adserà et al . (2020)] Enric Boix-Adserà, Benjamin L Edelman, and Siddhartha Jayanti. 2020. The multiplayer colonel blotto game. In Proceedings of the 21st ACM Conference on Economics and Computation . 47–48.
- 7[Borel(1921)] Emile Borel. 1921. La théorie du jeu et les équations intégralesa noyau symétrique. Comptes rendus de l’Académie des Sciences 173, 1304-1308 (1921), 58.
- 8[Golman and Page(2009)] Russell Golman and Scott E Page. 2009. General Blotto: games of allocative strategic mismatch. Public Choice 138, 3 (2009), 279–299.